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Q. $\displaystyle\lim _{x \rightarrow \infty}\left[\frac{x^{2}+x+3}{x^{2}-x+2}\right]^{X}$ is equal to

TS EAMCET 2016

Solution:

Given, $\displaystyle\lim _{x \rightarrow \infty} \left[\frac{x^{2}+x+3}{x^{2}-x+2}\right]^{x}$
$=\displaystyle\lim _{x \rightarrow \infty}\left[1-1+\frac{x^{2}+x+3}{x^{2}-x+2}\right]^{x}$
$=\displaystyle\lim _{x \rightarrow \infty}\left[1+\frac{x^{2}+x+3-x^{2}+x-2}{x^{2}-x+2}\right]^{x} $
$=\displaystyle\lim _{x \rightarrow \infty}\left[1+\frac{2 x+1}{x^{2}-x+2}\right]^{x} $
$=e^{\displaystyle\lim _{x \rightarrow \infty} \frac{2 x+1}{x^{2}-x+2} \cdot(x)} $
$\left[\therefore \displaystyle\lim _{x \rightarrow \infty}[1+f(x)]^{g(x)}=e^{\displaystyle\lim _{x \rightarrow \infty} f(x) \cdot g(x)}\right] $
$=e^{\displaystyle\lim _{x \rightarrow \infty} \frac{2 x^{2}+x}{x^{2}-x+2}}=e^{2}$